# ---
# title: 975. Odd Even Jump
# id: problem975
# author: Tian Jun
# date: 2020-10-31
# difficulty: Hard
# categories: Dynamic Programming, Stack, Ordered Map
# link: <https://leetcode.com/problems/odd-even-jump/description/>
# hidden: true
# ---
# 
# You are given an integer array `A`. From some starting index, you can make a
# series of jumps. The (1st, 3rd, 5th, ...) jumps in the series are called
# **odd-numbered jumps** , and the (2nd, 4th, 6th, ...) jumps in the series are
# called **even-numbered jumps**. Note that the **jumps** are numbered, not the
# indices.
# 
# You may jump forward from index `i` to index `j` (with `i < j`) in the
# following way:
# 
#   * During **odd-numbered jumps** (i.e., jumps 1, 3, 5, ...), you jump to the index `j` such that `A[i] <= A[j]` and `A[j]` is the smallest possible value. If there are multiple such indices `j`, you can only jump to the **smallest** such index `j`.
#   * During **even-numbered jumps** (i.e., jumps 2, 4, 6, ...), you jump to the index `j` such that `A[i] >= A[j]` and `A[j]` is the largest possible value. If there are multiple such indices `j`, you can only jump to the **smallest** such index `j`.
#   * It may be the case that for some index `i`, there are no legal jumps.
# 
# A starting index is **good** if, starting from that index, you can reach the
# end of the array (index `A.length - 1`) by jumping some number of times
# (possibly 0 or more than once).
# 
# Return _the number of **good** starting indices_.
# 
# 
# 
# **Example 1:**
# 
#     
#     
#     Input: A = [10,13,12,14,15]
#     Output: 2
#     Explanation: 
#     From starting index i = 0, we can make our 1st jump to i = 2 (since A[2] is the smallest among A[1], A[2], A[3],
#     A[4] that is greater or equal to A[0]), then we cannot jump any more.
#     From starting index i = 1 and i = 2, we can make our 1st jump to i = 3, then we cannot jump any more.
#     From starting index i = 3, we can make our 1st jump to i = 4, so we have reached the end.
#     From starting index i = 4, we have reached the end already.
#     In total, there are 2 different starting indices i = 3 and i = 4, where we can reach the end with some number of
#     jumps.
#     
# 
# **Example 2:**
# 
#     
#     
#     Input: A = [2,3,1,1,4]
#     Output: 3
#     Explanation: 
#     From starting index i = 0, we make jumps to i = 1, i = 2, i = 3:
#     
#     During our 1st jump (odd-numbered), we first jump to i = 1 because A[1] is the smallest value in [A[1], A[2],
#     A[3], A[4]] that is greater than or equal to A[0].
#     
#     During our 2nd jump (even-numbered), we jump from i = 1 to i = 2 because A[2] is the largest value in [A[2], A[3],
#     A[4]] that is less than or equal to A[1]. A[3] is also the largest value, but 2 is a smaller index, so we can
#     only jump to i = 2 and not i = 3
#     
#     During our 3rd jump (odd-numbered), we jump from i = 2 to i = 3 because A[3] is the smallest value in [A[3], A[4]]
#     that is greater than or equal to A[2].
#     
#     We can't jump from i = 3 to i = 4, so the starting index i = 0 is not good.
#     
#     In a similar manner, we can deduce that:
#     From starting index i = 1, we jump to i = 4, so we reach the end.
#     From starting index i = 2, we jump to i = 3, and then we can't jump anymore.
#     From starting index i = 3, we jump to i = 4, so we reach the end.
#     From starting index i = 4, we are already at the end.
#     In total, there are 3 different starting indices i = 1, i = 3, and i = 4, where we can reach the end with some
#     number of jumps.
#     
# 
# **Example 3:**
# 
#     
#     
#     Input: A = [5,1,3,4,2]
#     Output: 3
#     Explanation: 
#     We can reach the end from starting indices 1, 2, and 4.
#     
# 
# 
# 
# **Constraints:**
# 
#   * `1 <= A.length <= 2 * 104`
#   * `0 <= A[i] < 105`
# 
# 
## @lc code=start
using LeetCode

## add your code here:
## @lc code=end
